Let A and B be two distinct points on the line $$L : \dfrac{x - 6}{3} = \dfrac{y - 7}{2} = \dfrac{z - 7}{-2}$$. Both A and B are at a distance $$2\sqrt{17}$$ from the foot of perpendicular drawn from the point $$(1, 2, 3)$$ on the line L. If O is the origin, then $$\vec{OA} \cdot \vec{OB}$$ is equal to:

The given line $$L$$ can be written in parametric form as
$$x = 6 + 3t,\; y = 7 + 2t,\; z = 7 - 2t$$
so its direction vector is $$\mathbf{d} = \langle 3,\,2,\,-2\rangle$$.

Let the point $$P(1,\,2,\,3)$$ drop a perpendicular to $$L$$ at the foot $$H(6+3t,\,7+2t,\,7-2t)$$.
For the foot of the perpendicular, $$\overrightarrow{PH} \cdot \mathbf{d}=0$$.

Compute $$\overrightarrow{PH}$$:
$$\overrightarrow{PH}= \langle 6+3t-1,\; 7+2t-2,\; 7-2t-3\rangle =\langle 5+3t,\; 5+2t,\; 4-2t\rangle.$$

Form the dot product and set it to zero:
$$(5+3t)\cdot 3 + (5+2t)\cdot 2 + (4-2t)\cdot (-2) = 0.$$

Simplify:
$$3(5+3t) + 2(5+2t) -2(4-2t)=0$$
$$15+9t + 10+4t -8+4t = 0$$
$$17 + 17t = 0$$
$$\Rightarrow \; t = -1.$$

Thus
$$H = \bigl(6+3(-1),\,7+2(-1),\,7-2(-1)\bigr) = (3,\,5,\,9).$$

The magnitude of the direction vector is
$$|\mathbf{d}|=\sqrt{3^{2}+2^{2}+(-2)^{2}}=\sqrt{9+4+4}=\sqrt{17}.$$

Two points $$A$$ and $$B$$ on $$L$$ are each at distance $$2\sqrt{17}$$ from $$H$$.
If their parameters are $$t_1$$ and $$t_2$$ then
$$|t_1+1|\cdot|\mathbf{d}| = 2\sqrt{17},\qquad |t_2+1|\cdot|\mathbf{d}| = 2\sqrt{17}.$$

Since $$|\mathbf{d}|=\sqrt{17}$$, this gives
$$|t_1+1| = 2,\quad |t_2+1| = 2 \; \Longrightarrow\; t_1 = -3,\; t_2 = 1.$$

Coordinates of the two points:

$$t=-3 \Longrightarrow A(-3,\,1,\,13).$$

$$t=1 \Longrightarrow B(9,\,9,\,5).$$

Vectors from the origin are therefore
$$\vec{OA}=\langle -3,\,1,\,13\rangle,\qquad \vec{OB}=\langle 9,\,9,\,5\rangle.$$

The required dot product is
$$\vec{OA}\cdot\vec{OB} = (-3)(9) + (1)(9) + (13)(5) = -27 + 9 + 65 = 47.$$

Hence $$\vec{OA}\cdot\vec{OB}=47$$.
Option B is correct.